Solutions to the Homework III
I601 Logic and Discrete Mathematics
1. Show that (p → q) ∨ (p → r) and p → (q ∨ r) are logically equivalent. Try the
sequent calculus inference rules.
S o l u t i o n. We need to show provability of both (p → q) ∨ (p → r) → (p → (q ∨ r)) and
(p → (q ∨ r)) → (p → q) ∨ (p → r).
p, q ` q, r p ` p, q, r
p, r ` q, r p, ` p, q, r
(→L)
(→L)
p, (p→q) ` q, r
p, (p→r) ` q, r
(∨L)
(p → q)∨(p → r), p ` q, r
(∨R)
(p → q) ∨ (p → r), p ` q∨r
(→R)
(p → q) ∨ (p → r) ` p→(q ∨ r)
(→R)
` (p → q) ∨ (p → r)→(p → (q ∨ r))
and
q, p ` q, r r, p ` q, r
(∨L)
q∨r, p ` q, r
p ` p, q, r
(→L)
p→(q ∨ r), p ` q, r
(→R)
p → (q ∨ r), p ` q, p→r
(→R)
p → (q ∨ r) ` p→q, p → r
(∨R)
p → (q ∨ r) ` (p → q)∨(p → r)
(→R)
` (p → (q ∨ r))→(p → q) ∨ (p → r)
2.
Show that (p ∨ q) ∧ (p ∨ r) → (q ∨ r) is not tautology.
S o l u t i o n.
p ` q, r
p, r ` q, r
(∨L)
p, p∨r ` q, r
q, p ∨ r ` q, r
p∨q, p ∨ r ` q, r
(∧L)
(p ∨ q)∧(p ∨ r) ` q, r
(∨R)
(p ∨ q) ∧ (p ∨ r) ` q∨r
(→R)
` (p ∨ q) ∧ (p ∨ r)→(q ∨ r)
(∨L)
This formula is false for an interpretation that assigns p true and both q and r false:
(T ∨ F) ∧ (T ∨ F) → (F ∨ F) ≡ T ∧ T → F ≡ T → F ≡ F.
3. Determine the truth values of the statements ∀x∃y(xy = 1) and ∃x∀y(x 6 y 2 ) if
the domain for the variables consists of
a) the nonzero real numbers;
b) the nonzero integers;
c) the positive real numbers.
S o l u t i o n. Let P = ∀x∃y(xy = 1) and Q = ∃x∀y(x 6 y 2 ).
a) The domain of the function y = 1/x is all non-zero real numbers, therefore P = T.
Any perfect square is a non-negative number, so −1 6 y 2 for every y, and thus, Q = T.
b) y cannot be an integer for any x > 1 and P = F, but Q = T as for the case a).
c) P = T for same reason as in case a).
Whatever x > 0 we choose, the inequality x 6 y 2 is invalid for y <
x = 4, then x 6 y 2 is not satisfied for 0 < y < 2. So, Q = F.
√
x. For example, if